Consider two function $f(x)$ and $g(x)$ defined on [0,1] and $f(0) = g(0) = 0$ and $f(1) = g(1) = 1$. $f(x)$ is concave and weakly increasing, while $g(x)$ is convex and weakly increasing. Define the semi-elasticity of a function $h(x)$ as $\frac{h'(x)}{h(x)}$. Can you prove that the semi-elasticity of $g$ is greater than the semi-elasticity of $f$ for all $x \in (0,1)$?
It feels like this should be true. Working through a simple case where $f(x) = x^a$ and $g(x) = x^b$ for $0 < a < 1 < b$ we get $\frac{b}{x} > \frac{a}{x}$ for the semi-elasticities. But I'm not sure for the more general case. Any ideas?
Yes, it is true. Let $x\in (0, 1)$ and let $\phi(t)$ be the linear function such that $\phi(0)=0$ and $\phi(x)=f(x)$. The graph of $f$ will lie above the graph of $\phi$ on $(0, x)$, since $f$ is concave, but then a drawing shows that necessarily $f'(x)\leq \phi'(x)={f(x)\over x}$. In other words, ${f'(x)\over f(x)}\leq {1\over x}$.
Similarly, we get $g'(x)\geq {g(x)\over x}$, and hence ${g'(x)\over g(x)}\geq {1\over x}$, by comparing the graph of the convex function $g$ with the straight line through $(0, 0)$ and $(x, g(x))$.
Note that the assumption $f(1)=g(1)=1$ was not needed. This can also be seen by noting that ${h'(x)\over h(x)}$ is unchanged under vertical scaling, as $h(x)$ and $h'(x)$ will be multiplied by the same constant.